Answer:
a. Tm = 0.3192min.
b. MRR = 396.91mm^{3}/s.
Explanation:
Given the following data;
Drill diameter, D = 12.7mm
Depth, L = 60mm
Cutting speed, V = 25m/min = 25,000m
Feed, F = 0.30mm/rev
To find the cutting time;
Cutting time, Tm =?
[tex]Tm = \frac{L}{Fr}[/tex] .......eqn 1
We would first solve for the feed rate (F);
[tex]Fr = NF[/tex] .......eqn 2
But we need to find the rotational speed (N);
[tex]N= \frac{V}{\pi *D}[/tex]
[tex]N= \frac{25000}{3.142*12.7}[/tex]
[tex]N= \frac{25000}{39.90}[/tex]
N = 626.57rev/min.
Substiting N into eqn 2;
[tex]Fr = NF[/tex]
Fr = 626.57 * 0.30
Fr = 187.97mm/min.
Substiting F into eqn 1;
[tex]Tm = \frac{L}{Fr}[/tex]
[tex]Tm = \frac{60}{187.97}[/tex]
Tm = 0.3192min.
Therefore, the cutting time is 0.3192 minutes.
For the material removal rate (MRR);
[tex]MRR = \frac{\pi *D^{2}Fr}{4}[/tex]
[tex]MRR = \frac{3.142*12.7^{2}*187.97}{4}[/tex]
[tex]MRR = \frac{3.142*161.29*187.97}{4}[/tex]
[tex]MRR = \frac{95258.16}{4}[/tex]
[tex]MRR = 23814.54mm^{3}/min[/tex]
Time in seconds, we divide by 60;
MRR = 23814.54/60 =396.91mm^{3}/s.
Therefore, the material removal rate (MRR) is 396.91mm^{3}/s.
The boy in the wagon begins throwing bricks out of the wagon to simulate rocket propulsion. The wagon begins at rest, and the boy throws three bricks. The boy’s weight is 80lbs, and the weight of the wago 20 lbs. The bricks weigh 10 lbs each, and he throws them with a horizontal velocity of 10 ft/s relative to the wagon. Neglect horizontal forces on the wagon's wheels.
A) What velocity does the boy attain if he throws the bricks one at a time?
B) What velocity does the boy attain if he throws all three bricks at once?
C) Compare the results of parts (a) and (b). What does this suggest about rocket propulsion? Why are these results different?
Q:What velocity does the boy attain if he throws the bricks one at a time?
Answer:Linear velocity since it moves back and firth and does not rotate like angular velocity.
The system is stimulated, via the voltage source, with a pulse of height 2 and width 4 s. Determine the voltage across the resistor.
Answer:
Voltage across resistor = 2 v
Explanation:
Given data
pulse height = 2 v
pulse width = 4s
calculate voltage across resistor ( the free hand sketch attached below explains more )
pulse height is also = amplitude of voltage ) = 2v
The voltage across the resistor = 2v Since the voltage from the source of the circuit is equal to the amplitude voltage in the circuit ( assuming no loss of voltage )
also the graphical representation of the problem is attached below
Be-16 a garbage dumping placard must be prominently posted on boats longer than what size?
Answer:
26 feet and longer boats that have garbage dumping placard must be prominently posted and the boats which are 40 feet and longer must have the written waste management plan.
A 60-Hz 220-V-rms source supplies power to a load consisting of a resistance in series with an inductance. The real power is 1500 W, and the apparent power is 4600 VA.
a. Determine the value of the resistance.
b. Determine the value of the inductance.
Answer:
(a) The value of the resistance is 3.431 Ω
(b) The value of the inductance is 0.0264 H
Explanation:
Given;
frequency of the source, f = 60 Hz
rms voltage, V-rms = 220 V
real power, Pr = 1500 W
apparent power, Pa = 4600 VA
(a). Determine the value of the resistance
[tex]P_r = I_{rms}^2R[/tex]
where;
R is resistance
[tex]I_{rms} = \frac{Apparent \ Power}{V_{rms}} \\\\I_{rms} = \frac{P_a}{V_{rms}}\\\\I_{rms}= \frac{4600}{220} \\\\I_{rms}= 20.91 \ A[/tex]
Resistance is calculated as;
[tex]R = \frac{P_r}{I_{rms}^2} \\\\R = \frac{1500}{(20.91)^2} \\\\R = 3.431 \ ohms[/tex]
(b). Determine the value of the inductance.
[tex]Q_L = I_{rms}^2 X_L[/tex]
where;
[tex]Q_L[/tex] is reactive power
[tex]X_L[/tex] is inductive reactance
[tex]Apparent \ power = \sqrt{Q_L^2 + P_r^2} \\\\P_a^2 = Q_L^2 + P_r^2\\\\Q_L^2 = P_a^2 - P_r^2\\\\Q_L^2 = 4600^2 - 1500^2\\\\Q_L^2 = 18910000\\\\Q_L = \sqrt{18910000}\\\\Q_L = 4348.56 \ VA[/tex]
inductive reactance is calculated as;
[tex]X_L = \frac{Q_L}{I_{rms}^2} \\\\X_L = \frac{4348.56}{(20.91)^2} \\\\X_L = 9.95 \ ohms[/tex]
inductance is calculated as;
[tex]X_L = \omega L\\\\X_L = 2\pi f L\\\\L = \frac{X_L}{2\pi f} \\\\L = \frac{9.95}{2\pi *60} \\\\L = 0.0264 \ H\\\\L = 26.4 \ mH[/tex]
A car travels from A, due north to a town B 4 km away. It then travels due east until it arrives town C 5 km from B. determine the distance of town C from A
Answer:
A to C = 6.4 km
Explanation:
A to B = 4 km
B to C = 5 km
A to C = using pythagorean theorem
a² + b² = c²
a = A to B = 4
b = B to C = 5
c = A to C
c² = 4² + 5²
c = 6.4 km (A to C)
According to the scenario, the distance between town C from town A is found to be 6.40 Km.
Which background does this question depend on?The background that this question depends on is known as the direction-based question. These types of questions completely depend on the distance of moving bodies like cars, persons, or any other objects as well with respect to the initial position.
According to the question,
The distance between town A to town B = 4 km.
The distance between town B to town C = 5 km.
Now, according to the Pythagoras theorem, the distance between town C to town A is as follows:
[tex]AC^2[/tex] = [tex]AB^2 +BC^2[/tex].
[tex]AC^2[/tex] = [tex]4^2+5^2[/tex]
[tex]AC^2[/tex] = 16 + 25 = 41.
AC = √41 = 6.40 km.
Therefore, the distance between town C from town A is found to be 6.40 Km.
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A cylindrical tank is 50 inches long, has a diameter of 16 inches and contains 1.65 lbm of water. Calculate the density of water in Ibm per cubic feet.
Answer:
0.285 lbm per cubic feet
Explanation:
length of tank = 50 inches = 50/12 feet = 4.17 feet
diameter of tank = 16 inches = 16/12 feet = 1.33 feet
weight of water = 1.65 lbm
density of water = ?
We know that the density of a substance is given as
ρ = w/v
where ρ is the density in Ibm per cubic feet
w is the weight in lbm
v is the volume in cubic feet
Volume of a cylinder = [tex]\frac{\pi d^{2} l}{4}[/tex]
where d is the diameter
l is the length
volume = [tex]\frac{3.142*1.33^{2}* 4.17}{4}[/tex] = 5.79 cubic feet
Therefore, the density of water will be
ρ = w/v = 1.65/5.79 = 0.285 lbm per cubic feet
The density of water will be "0.285 lbm per cubic feet".
Density:According to the question :
Length of tank = 50 inches or,
= [tex]\frac{50}{12}[/tex]
= 4.17 feet
Diameter of tank = 16 inches or,
= [tex]\frac{16}{12}[/tex]
= 1.33 feet
Weight of water = 1.65 lbm
We know the formula,
Volume of cylinder,
= [tex]\frac{\pi d^2 l}{2}[/tex]
By substituting the values, we het
= [tex]\frac{3.142\times (1.33)^2\times 4.17}{4}[/tex]
= 5.79 cubic feet
hence,
The density of water will be:
→ ρ = [tex]\frac{w}{v}[/tex]
By substituting the values,
= [tex]\frac{1.65}{5.79}[/tex]
= 0.285 lbm/cubic feet
Thus the above answer is correct.
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If you see a red, a green, and a white light on another boat, what does this tell you?
A boat is approaching you head on.
The red and green lights are sidelights that are positioned on the port side (red) (left as facing the bow) and starboard (green) (right as facing the bow) side of the boat. Various white lights are required depending on the size of the boat, but generally, a white masthead light and stern light are required. See the US Coast Guard site in the link below for more specific information.
Hope this helps
The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the ______________ speed and varies with weight
Answer:
Maneuvering speed.
Explanation:
The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the maneuvering speed and varies with weight.
In aeronautical engineering, the maneuvering speed (Va) of an aircraft such as an aeroplane, helicopter, or jet is an airspeed limitation which is mainly selected by an aircraft designer.
Generally, at speeds higher or greater than the manoeuvring speed, aircraft pilots are advised not to attempt a full deflection of any flight control surface because it's capable of resulting in a damage to the structure of an aircraft.
If you're a pilot, to find the maneuvering speed of an aircraft, you should look at the flight manual of the aircraft or on the cockpit placard in the aircraft. The maneuvering speed of an aircraft is a calibrated speed and should not be exceeded by any pilot.
A 208-V, 10hp, four pole, 60 Hz, Y-connected
induction motor has a full-load slip of 5 percent
1. What is the synchronous speed of this motor?
2. What is the rotor speed of this motor at rated
load?
3. What is the rotor frequency of this motor at
rated load?
4. What is the shaft torque of this motor at rated
load?
A wall 0.12 m thick having a thermal diffusivity of 1.5 × 10-6 m2/s is initially at a uniform temperature of 97°C. Suddenly one face is lowered to a temperature of 20°C, while the other face is perfectly insulated. Use the explicit finite-difference technique with space and time increments of 30 mm and 300 s to determine the temperature distribution at at 45 minutes.
Answer:
at t = 45 s :
To = 61.7⁰c, T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C
Explanation:
Wall thickness = 0.12 m
thermal diffusivity = 1.5 * 10^-6 m^2/s
Δt ( time increment ) = 300 s
Δ x = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )
using the explicit finite-difference technique
Detailed solution is attached below
If you are involved in a collision where there is injury, you must report the incident within .......
48 hours
3 days
24 hours
72 hours
Answer:
24 hours
Explanation:
you must exchange insurance details after a collision if someone is injured. Otherwise you must report the collision to us as soon as possible (and no later than 24 hours). Although you must report such a collision straight away you should always seek medical help in the first instance.
Write torsion equation and explain the importance of each components.
Andy and Wendy both wear glasses. On a cold winter day, Andy comes from the cold outside and enters the warm house, while Wendy leaves the house and goes outside. Are Wendy's glasses more likely to be fogged than Andy's? Explain.
Answer:
Andy's glasses are more likely to fog up because the cold air is being conducted by the hot air somthing like that
A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in the core is 1.8 m and the joints are equivalent to an airgap of 0.1 mm. The value of the magnetic field strength for 1.1 T in the core is 400 A/m, the corresponding core loss is 1.7 W/kg at 50 Hz and the density of the core is 7800 kg/m3. If the maximum value of the flux density is to be 1.1 T when a p.D. Of 2200 V at 50 Hz is applied to the primary, calculate: a. The cross-sectional area of the core; b. The secondary voltage on no load; c. The primary current and power factor on no load.
Answer:
a) cross sectional area of the core = 0.0187 m²
b) The secondary voltage on no-load = 413 V
c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.
Explanation:
See attached solution.
The cars of a roller-coaster ride have a speed of 30 km / h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 18 m, and all six cars have the same mass.
Answer:
Explanation:
30 we know that radius is 18 and the circumference is 36pi and the time to go around is is 36pi/30=1.2pi≈3.76991118
A small still is separating propane and butane at 135 °C, and initially contains 10 kg moles of a mixture whose composition is x = 0.3 (x = mole fraction butane). Additional mixture (x = 0.3) is fed at the rate of 5 kg mole/hr. The total volume of the liquid in the still is constant, and the concentration of the vapor from the still (xp) is related to x, as follows: Xp = How long will it take for X, to change from 0.3 to 0.35.
Answer:
Hello the needed relation is missing below is the required relation
[tex]X_{p} = \frac{x_{s} }{1+x_{s} }[/tex] composition : propane = 0.70, butane = 0.3
Answer : ≈ 5.75 hrs
Explanation:
Applying the data given in regards to the material balance
Butane balance input into the still = 5 mole feed/hr | 0.30 mol butane/molfeed
since the total volume of the liquid in the still is constant
The output from the still is = 5mol condensed/hr | x[tex]_{p}[/tex] mol butane/mol condensed
unsteady state equation = [tex]\frac{dx_{s} }{dt}[/tex] = 0.15 - [tex]0.5X_{p}[/tex]
note : to reduce the equation a single dependent variable we have to substitute for [tex]x_{p}[/tex]
[tex]\frac{dx_{s} }{dt}[/tex] [tex]= 0.15 + x_{s} / 1 + (0.5)x_{s}[/tex]
In order to find the time it will take for X to change from 0.3 to 0.35
integrate the above equation using the limits : t = 0, x[tex]_{s}[/tex] = 0.3 and t = Ф,
x[tex]_{s} = 0.35[/tex]
= [tex][ - (x_{s} /0.35 - (1/(0.35)^2)* In(0.15 - 0.35x_{s} ) ]_{0.3} ^{0.35}[/tex]
hence t = Ф ≈ 5.75 hrs
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
Answer: 164.2253 MPa
Explanation:
First we find the half internal crack which is = length of surface crack /2
so α = 8.6 /2 = 4.3mm ( 4.3×10⁻³m )
Now we find the dimensionless parameter using the critical stress crack propagation equation
∝ = K / Y√πα
stress level ∝ = 112Mpa
fracture toughness K = 26Mpa
dimensionless parameter Y = ?
SO working the formula
Y = K / ∝√πα
Y = 26 / 112 (√π × 4.3× 10⁻³)
Y = 1.9973
We are asked to find stress level for internal crack length of 4m
so half internal crack is = length of surface crack /2
4/2 = 2mm ( 2 × 10⁻³)
from the previous formula critical stress crack propagation equation
∝ = K / Y√πα
∝ = 26 / 1.9973 √(π × 2 × 10⁻³)
∝ = 164.2253 Mpa
Which statement about tensile stress is true? A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object. B. Forces that act perpendicular to the surface and squeeze an object exert a tensile stress on the object. C. Forces that act parallel to the surface exert a tensile stress on the object. D. Forces that decrease the length of the material exert a tensile stress on the object.
Answer:
A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.
Explanation:
Tensile stress is due to tension forces on a material. Tensile force acts perpendicularly away from the surface of the substance. The pull on the material due to the tensile force exerts tensile stress on the material, that tends to pull the material apart. The magnitude of the tensile stress is given as
σ = [tex]\frac{P}{A}[/tex]
where σ is the tensile stress
P is the tensile force pulling the material apart
A is the cross-sectional area through which the tensile force acts perpendicularly.
Determine the normal stress in a ball, which has an outside diameter of 160 mm and a wall thickness of 3.8 mm, when the ball is inflated to a gage pressure of 78 kPa.
Answer:
The normal stress is 0.7821 MPa
Explanation:
The external diameter D = 160 mm
The thickness t = 3.8 mm = 3.8 x 10^-3 m
gauge pressure P = 78 kPa = 78 x 10^3 Pa
The maximum shear stress τmax = ?
The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm
The internal radius of the shell r = R - t
==> 80 - 3.8 = 76.2 mm
Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm
==> d = 152.4 x 10^-3 m
The normal stress σ = [tex]\frac{Pd}{4t}[/tex] = [tex]\frac{78*10^{3}*152.4*10^{-3} }{4*3.8*10^{-3} }[/tex] = 782052.63 Pa
==> σ = 0.7821 MPa
The screw of shaft straightener exerts a load of 30 as shown in Figure . The screw is square threaded of outside diameter 75 mm and 6 mm pitch.
force required at the rim of a 300mm diameter hand wheel, if there is a collar
bearing of 50 mm mean diameter provided in the arrangement to exert axial
load. Assume the coefficient of friction for the collar as 0.2.
Answer:
See calculation below
Explanation:
Given:
W = 30 kN = 30x10³ N
d = 75 mm
p = 6 mm
D = 300 mm
μ = tan Φ = 0.2
1. Force required at the rim of handwheel
Let P₁ = Force required at the rim of handwheel
Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm
Mean diameter of screw: *d = [tex]\frac{do + dc}{2}[/tex] = (75 + 69) / 2 = 72 mm
and
tan α = p / πd = 6 / (π x 72) = 0.0265
∴ Torque required to overcome friction at he threads is T = P x d/2
T = W tan (α + Ф) d/2
T = [tex]W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}[/tex]
T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2
T = 245,400 N-mm
We know that the torque required at the rim of handwheel (T)
245,400 = P1 x D/2 = P1 x (300/2) = 150 P1
P1 = 245,400 / 150
P1 = 1636 N
2. Maximum compressive stress in the screw
30x10³
Qc = W / Ac = -------------- = 8.02 N/mm²
π/4 * 69²
Qc = 8.02 MPa
Bearing pressure on the threads (we know that number of threads in contact with the nut)
n = height of nut / pitch of threads = 150 / 6 = 25 threads
thickness of threads, t = p/2 = 6/2 = 3 mm
bearing pressure on the threads = Pb = W / (π d t n)
Pb = 30 x 10³ / (π * 72 * 3 * 25)
Pb = 1.77 N/mm²
Max shear stress on the threads = τ = 16 T / (π dc³)
τ = (16 * 245,400) / ( π * 69³ )
τ = 3.8 M/mm²
*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72
∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))
τmax = 5.5 Mpa
3. efficiency of the straightener
To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm
∴Efficiency of the straightener is η = To / T = 28,620 / 245,400
η = 0.116 or 11.6%
By assuming the coefficient of friction for the collar as 0.2. efficiency of straightner is 11.6%.
What is Wheel?Wheels are circular frames or disks that are mounted on machines or vehicles and are designed to rotate around an axis.
Given:
W = 30 kN = 30x10³ N
d = 75 mm
p = 6 mm
D = 300 mm
μ = tan Φ = 0.2
1. Force required at the rim of handwheel :
Let P₁ = Force required at the rim of handwheel
Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm
Mean diameter of screw: *d = = (75 + 69) / 2 = 72 mm and
tan α = p / πd = 6 / (π x 72) = 0.0265
Torque required to overcome friction at he threads is T = P x d/2
T = W tan (α + Ф) d/2
T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2
T = 245,400 N-mm
We know that the torque required at the rim of handwheel (T)
245,400 = P1 x D/2 = P1 x (300/2) = 150 P1
P1 = 245,400 / 150
P1 = 1636 N
2. Maximum compressive stress in the screw :
30x10³
Qc = W / Ac = -------------- = 8.02 N/mm²
π/4 * 69²
Qc = 8.02 MPa
Bearing pressure on the threads (we know that number of threads in contact with the nut)
n = height of nut / pitch of threads = 150 / 6 = 25 threads
thickness of threads, t = p/2 = 6/2 = 3 mm
bearing pressure on the threads = Pb = W / (π d t n)
Pb = 30 x 10³ / (π * 72 * 3 * 25)
Pb = 1.77 N/mm²
Max shear stress on the threads = τ = 16 T / (π dc³)
τ = (16 * 245,400) / ( π * 69³ )
τ = 3.8 M/mm²
*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72
max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))
τmax = 5.5 Mpa
Therefore, efficiency of the straightener :
To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm
Efficiency of the straightener is η = To / T = 28,620 / 245,400
η = 0.116 or 11.6%
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A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one- dimensional. Two thermocouples are embedded in each sample some distance (L) apart, and a differential thermometer reads the temperature drop (Delta T) across this distance along each sample. When steady-state operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, rectangular samples (5 cm Times 5 cm on the side exposed to the heater and 10 cm long) are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V, and both differential thermometers read a temperature difference of 15 degree C. Determine the thermal conductivity of the sample.
Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = VI = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C
An inventor claims to have developed a device requiring no work or heat transfer input yet able to produce hot and cold air streams at steady state.
a. True
b. False
Answer:
a. True
Explanation:
Apply the principle of conservation of mass.
and also the expression for the steady flow energy equation.
kinetic and potential energy effects can be neglected.
The given statement by the inventor who is claiming the development of a device that requires no work or heat transfer input yet is able to produce hot and cold air streams at a steady state is definitely false.
What is heat transfer?Heat transfer may be characterized as a type of process which involves the migration of heat from one object or component to another by numerous mechanisms like conduction, convection, and/or radiation.
The process of heat transfer may occur where there is a temperature difference between two objects exist. It significantly utilizes the mechanism of exchanging thermal energy between two or more physical systems.
According to the concept of physics, no object or thing has the ability to perform its function without the utilization of any source of heat or energy. Then, how it is possible for that device to produce hot and cold air streams at a steady state.
Therefore, the given statement by the inventor is absolutely false.
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In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very different register transfers to take place. Why does this not happen
Answer:
no need for that
Explanation:
they are not the same at all
4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD pivoted at C. A pin at end D of the rod fits into one of several holes drilled in the edge of the lid. For α 5 50°, determine (a) the magnitude of the force exerted by rod CD, (b) the reactions at the hinges. Assume that the hinge at B does not exert any axial thrust. Beer, Ferdinand. Vector Mechanics for Engineers: Statics (p. 219). McGraw-Hill Higher Education. Kindle Edition.
Answer:
(a) The magnitude of force is 116.6 lb, as exerted by the rod CD
(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.
Explanation:
Step by step working is shown in the images attached herewith.
For this given system, the coordinates are the following:
A(0, 0, 0)
B(26, 0, 0)
And the value of angle alpha is 20.95°
Hope that answers the question, have a great day!
A quartz window of thickness L serves as a viewing port in a furnace used for annealing steel. The inner surface (x=0) of the window is irradiated with a uniform heat flux q''o due to emission from hot gases in the furnace. A fraction, β, of this radiation may be assumed to be absorbed at the inner surface, while the remaining radiation is partially absorbed as it passes through the quartz. The volumetric heat generation due to this absorption may be described by an expression of the form q˙(x)=(1−β)q''oα^e−αx where α is the absorption coefficient of the quartz. Convection heat transfer occurs from the outer surface (x=L) of the window to ambient air at T[infinity] and is characterized by the convection coefficient h. Convection and radiation emission from the inner surface may be neglected, along with radiation emission from the outer surface.
Required:
Determine the temperature distribution in the quartz, expressing your result in terms of the foregoing parameters.
Answer:
Assuming steady state condition the temperature distribution is calculated as expressed in the attached solution below
Explanation:
Given data :
thickness : L , inner surface (x) : 0, uniform flux : q"o
fraction : β
volumetric heat generation : q˙(x)=(1−β)q''oα^e−αx
determine the temperature distribution in the quartz
attached below is the detailed solution
Match the words in the left column to the appropriate blanks in the sentences on the right. Note that some words may be used more than once and some may not be used.
1. breaking
2. forming
3. positive
4. negative
5. twice
6. half
A. The reaction involves___five blue-blue and twenty blue-red bonds and then____twenty blue-red bonds. Enthalpies for bond breaking are always_____.
B. In the depicted reaction, both reactants and products are assumed to be in the gas phase. There are___as many molecules of in the products, delta S is___for this reaction
Answer:
A. Breaking; forming; positive.
B. Twice; half.
Explanation:
A. The reaction involves breaking five blue-blue and twenty blue-red bonds and then forming twenty blue-red bonds. Enthalpies for bond breaking are always positive.
B. In the depicted reaction, both reactants and products are assumed to be in the gas phase. There are twice as many molecules of gas in the products, delta S is half for this reaction.
Bond enthalpy refers to the amount of energy that is required to break a mole of a particular bond and it's an endothermic reaction, therefore it is always positive.
However, bond forming is always an exothermic reaction because energy is released, therefore, it is always negative. This simply means that, the breaking of a bond is an endothermic reaction (positive) while the formation of a bond is an exothermic reaction (negative).
Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at 11 MPa. Determine the heat transfer in kJ per kg of steam flowing, for the working fluid passing through the boiler and condenser and calculate the thermal efficiency.
Answer:
19
Explanation:
when you plus it it will give this
the 8kpa and 11mpa are plused
Who plays a role in the financial activities of a company?
O A. Just employees
O B. Just managers
O C. Only members of the finance and accounting department
O D. Everyone at the company, including managers and employees
Hey,
Who plays a role in the financial activities of a company?
O D. Everyone at the company, including managers and employees
Answer:
Everyone at the company, including managers and employees
Explanation:
You are the curator of a museum. The museum is running short of funds, so you decide to increase revenue. Should you increase or decrease the price of admission? Explain
Answer:
Explanation:
If the museum is running short of funds, and you decide to increase revenue. An increase or decrease in the price of admission into the museum depends on the following:
1. If demand for admission into the museum is elastic there are two possible outcomes
a. An increase in the price of admission leads to a decrease in the quantity demand of admission into the museum
b. A decrease in price of admission into the museum leads to an increase in the quantity demand of admission into the museum.
This follows the law of demand which states that "the higher the price, the lower the quantity demanded and the lower the price, the higher the quantity demanded".
2. If the demand for admission into the museum is inelastic, then an increase in price will lead to an increase in revenue of the museum.
Therefore, before the curator increase the price of admission into the museum, he should first determine the price elasticity of demand of the museum.
Cold water at 20 degrees C and 5000 kg/hr is to be heated by hot water supplied at 80 degrees C and 10,000 kg/hr. You select from a manufacturer's catalog a shell-and-tube heat exchanger (one shell with two tube passes) having a UA value of 11,600 W/K. Determine the hot water outlet temperature.
Answer:
59°C
Explanation:
Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)
and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)
Therefore the minimum and maximum heat capacities are:
Cmin = Cc = 5803.2(W/K)
Cmax = Ch = 11634.3(W/K)
The capacity ratio is:
Cr = Cmin / Cmax = 0.499 = 0.5
The maximum possible heat transfer rate is:
Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)
And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99
Given that from the appropriate graph in the handouts we can read = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)
Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C